13.07.2015 Views

Mathematical Methods for Physics and Engineering - Matematica.NET

Mathematical Methods for Physics and Engineering - Matematica.NET

Mathematical Methods for Physics and Engineering - Matematica.NET

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

11.8 DIVERGENCE THEOREM AND RELATED THEOREMSSince v has a singularity at the origin it is not differentiable there, i.e. ∇·v is notdefined there, but at all other points ∇ · v = 0, as required <strong>for</strong> an incompressiblefluid. There<strong>for</strong>e, from the divergence theorem, <strong>for</strong> any closed surface S 2 that doesnot enclose the origin we have∮ ∫v · dS = ∇ · v dV =0.S 2 VThus we see that the surface integral ∮ Sv · dS has value Q or zero dependingon whether or not S encloses the source. In order that the divergence theorem isvalid <strong>for</strong> all surfaces S, irrespective of whether they enclose the source, we write∇ · v = Qδ(r),where δ(r) is the three-dimensional Dirac delta function. The properties of thisfunction are discussed fully in chapter 13, but <strong>for</strong> the moment we note that it isdefined in such a way that∫Vδ(r − a) =0 <strong>for</strong>r ≠ a,{f(a) if a lies in Vf(r)δ(r − a) dV =0 otherwise<strong>for</strong> any well-behaved function f(r). There<strong>for</strong>e, <strong>for</strong> any volume V containing thesource at the origin, we have∫∫∇ · v dV = Q δ(r) dV = Q,Vwhich is consistent with ∮ Sv · dS = Q <strong>for</strong> a closed surface enclosing the source.Hence, by introducing the Dirac delta function the divergence theorem can bemade valid even <strong>for</strong> non-differentiable point sources.The generalisation to several sources <strong>and</strong> sinks is straight<strong>for</strong>ward. For example,if a source is located at r = a <strong>and</strong> a sink at r = b then the velocity field isv =<strong>and</strong> its divergence is given byV(r − a)Q (r − b)Q−4π|r − a|34π|r − b| 3∇ · v = Qδ(r − a) − Qδ(r − b).There<strong>for</strong>e, the integral ∮ Sv · dS has the value Q if S encloses the source, −Q ifS encloses the sink <strong>and</strong> 0 if S encloses neither the source nor sink or enclosesthem both. This analysis also applies to other physical systems – <strong>for</strong> example, inelectrostatics we can regard the sources <strong>and</strong> sinks as positive <strong>and</strong> negative pointcharges respectively <strong>and</strong> replace v by the electric field E.405

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!